3. A water balloon is thrown at a 30° angle above the horizontal from a high balcony. The balloon remains in the air for 2.50 seconds before landing 43.30 meters from the base of the balcony.
How fast was the water balloon traveling at the apex of its trajectory? How fast was the water balloon thrown from the balcony? Ignore air resistance. Answers 17.32 m/s, 20.00 m/s
How fast was the water balloon traveling at the apex of its trajectory? How fast was the water balloon thrown from the balcony? Ignore air resistance. Answers 17.32 m/s, 20.00 m/s
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OK, first thing first. You have two different motions:
a) vertical: S=So+VoT-(1/2)AT^2; where A is actually the acceleration of gravity. S is space (distance or height) So is initial space (starting point with respect to the ground or an "origin").
b) horizontal: X=V*T; X is just the horizontal distance.
Ok, next thing; whenever dealing with a problem of parabloic motion the first thing you do is take the initial speed and multiply that by sin(angle) or cos(angle) for vertical and horizontal speeds respectively and then put them into the equations I gave you before. The only thing that is truly common between these two simultaneous motions is the time it takes the object in question to travel from point a to b.
Here, you're being asked to calculate the initial velocity. I would go with the equation for the horizontal motion to find the horizontal speed. Keep in mind that the balcony is at some vertical height So that is not given and that the final height is S=0 because it hits the ground.
So, you have X=Vocos(30)*T. You have X which is the distance form the base so for the initial velocity plug and chug.
Now the Apex of the speed, I do believe that means the highest point in which case you would need to find the maximum height first. I hope you know calculus...derive equation (a) w/ respect to T and find the max by making it equal to zero. The time T will give you the time it took to get to the max height. Substitute that T into equation (a) (not derived) and find out the max height. At that time you and with the initial velocity you should be able to find the horizontal distance. At the very "top of the trajectory" gravity stops to influence the balloon for a brief instant so you can ignore the vertical speed. Now you're left with X=V*T.
Damn...I didn't actually think I was going to go through the whole problem. I got caught up in it....oops
a) vertical: S=So+VoT-(1/2)AT^2; where A is actually the acceleration of gravity. S is space (distance or height) So is initial space (starting point with respect to the ground or an "origin").
b) horizontal: X=V*T; X is just the horizontal distance.
Ok, next thing; whenever dealing with a problem of parabloic motion the first thing you do is take the initial speed and multiply that by sin(angle) or cos(angle) for vertical and horizontal speeds respectively and then put them into the equations I gave you before. The only thing that is truly common between these two simultaneous motions is the time it takes the object in question to travel from point a to b.
Here, you're being asked to calculate the initial velocity. I would go with the equation for the horizontal motion to find the horizontal speed. Keep in mind that the balcony is at some vertical height So that is not given and that the final height is S=0 because it hits the ground.
So, you have X=Vocos(30)*T. You have X which is the distance form the base so for the initial velocity plug and chug.
Now the Apex of the speed, I do believe that means the highest point in which case you would need to find the maximum height first. I hope you know calculus...derive equation (a) w/ respect to T and find the max by making it equal to zero. The time T will give you the time it took to get to the max height. Substitute that T into equation (a) (not derived) and find out the max height. At that time you and with the initial velocity you should be able to find the horizontal distance. At the very "top of the trajectory" gravity stops to influence the balloon for a brief instant so you can ignore the vertical speed. Now you're left with X=V*T.
Damn...I didn't actually think I was going to go through the whole problem. I got caught up in it....oops